3.3: Truth Tables- Conditional, Biconditional
A conditional is a logical compound statement in which a statement \(p\), called the hypothesis, implies a statement \(q\), called the conclusion.
A conditional is written as \(p \rightarrow q\) and is translated as "if \(p\), then \(q\)".
The English statement “If it is raining, then there are clouds is the sky” is a conditional statement. It makes sense because if the hypothesis “it is raining” is true, then the conclusion “there are clouds in the sky” must also be true.
Notice that the statement tells us nothing of what to expect if it is not raining; there might be clouds in the sky, or there might not. If the hypothesis is false, then the conclusion becomes irrelevant.
Suppose you order a team jersey online on Tuesday and want to receive it by Friday so you can wear it to Saturday’s game. The website says that if you pay for expedited shipping, you will receive the jersey by Friday. In what situation is the website telling a lie?
There are four possible outcomes:
- You pay for expedited shipping and receive the jersey by Friday
- You pay for expedited shipping and don’t receive the jersey by Friday
- You don’t pay for expedited shipping and receive the jersey by Friday
- You don’t pay for expedited shipping and don’t receive the jersey by Friday
Only one of these outcomes proves that the website was lying: the second outcome in which you pay for expedited shipping but don’t receive the jersey by Friday. The first outcome is exactly what was promised, so there’s no problem with that. The third outcome is not a lie because the website never said what would happen if you didn’t pay for expedited shipping; maybe the jersey would arrive by Friday whether you paid for expedited shipping or not. The fourth outcome is not a lie because, again, the website didn’t make any promises about when the jersey would arrive if you didn’t pay for expedited shipping.
It may seem strange that the third outcome in the previous example, in which the first part is false but the second part is true, is not a lie. Remember, though, that if the hypothesis is false, we cannot make any judgment about the conclusion. The website never said that paying for expedited shipping was the only way to receive the jersey by Friday.
A friend tells you “If you upload that picture to Facebook, you’ll lose your job.” Under what conditions can you say that your friend was wrong?
 In the last two cases, your friend didn’t say anything about what would happen if you didn’t upload the picture, so you can’t say that their statement was wrong. Even if you didn’t upload the picture and lost your job anyway, your friend never said that you were guaranteed to keep your job if you didn’t upload the picture; you might lose your job for missing a shift or punching your boss instead.
\(\begin{array}{|c|c|c|}
\hline p & q & p \rightarrow q \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
Again, if the hypothesis \(p\) is false, we cannot prove that the statement is a lie, so the result of the third and fourth rows is true.
Construct a truth table for the statement \((m \wedge \sim p) \rightarrow r\)
Solution
We start by constructing a truth table with 8 rows to cover all possible scenarios. Next, we can focus on the hypothesis, \(m \wedge \sim p\).
\(\begin{array}{|c|c|c|}
\hline m & p & r \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline
\end{array}\)
\(\begin{array}{|c|c|c|c|}
\hline m & p & r & \sim p \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
\(\begin{array}{|c|c|c|c|c|}
\hline m & p & r & \sim p & m \wedge \sim p \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}\)
Now we can create a column for the conditional. Because it can be confusing to keep track of all the Ts and \(\mathrm{Fs}\), why don't we copy the column for \(r\) to the right of the column for \(m \wedge \sim p\) ? This makes it a lot easier to read the conditional from left to right.
\(\begin{array}{|c|c|c|c|c|c|c|}
\hline m & p & r & \sim p & m \wedge \sim p & r & (m \wedge \sim p) \rightarrow r \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
When \(m\) is true, \(p\) is false, and \(r\) is false- -the fourth row of the table-then the hypothesis \(m \wedge \sim p\) will be true but the conclusion false, resulting in an invalid conditional; every other case gives a valid conditional.
If you want a real-life situation that could be modeled by \((m \wedge \sim p) \rightarrow r\), consider this: let \(m=\) we order meatballs, \(p=\) we order pasta, and \(r=\) Rob is happy. The statement \((m \wedge \sim p) \rightarrow r\) is "if we order meatballs and don't order pasta, then Rob is happy". If \(m\) is true (we order meatballs), \(p\) is false (we don't order pasta), and \(r\) is false (Rob is not happy), then the statement is false, because we satisfied the hypothesis but Rob did not satisfy the conclusion.
For any conditional, there are three related statements, the converse, the inverse, and the contrapositive.
The original conditional is \(\quad\) "if \(p,\) then \(q^{\prime \prime} \quad p \rightarrow q\)
The converse is \(\quad\) "if \(q,\) then \(p^{\prime \prime} \quad q \rightarrow p\)
The inverse is \(\quad\) "if not \(p,\) then not \(q^{\prime \prime} \quad \sim p \rightarrow \sim q\)
The contrapositive is "if not \(q,\) then not \(p^{\prime \prime} \quad \sim q \rightarrow \sim p\)
Consider again the conditional “If it is raining, then there are clouds in the sky.” It seems reasonable to assume that this is true.
The converse would be “If there are clouds in the sky, then it is raining.” This is not always true.
The inverse would be “If it is not raining, then there are not clouds in the sky.” Likewise, this is not always true.
The contrapositive would be “If there are not clouds in the sky, then it is not raining.” This statement is true, and is equivalent to the original conditional.
Looking at truth tables, we can see that the original conditional and the contrapositive are logically equivalent, and that the converse and inverse are logically equivalent.
A conditional statement and its contrapositive are logically equivalent.
The converse and inverse of a conditional statement are logically equivalent.
In other words, the original statement and the contrapositive must agree with each other; they must both be true, or they must both be false. Similarly, the converse and the inverse must agree with each other; they must both be true, or they must both be false.
We typically represent the conditional using the words, "if ..., then ...," but there are other ways this logical connective can be represented in English. Consider the conditional from Example 5: "If it is raining, then there are clouds in the sky." We could equivalently write, "It is raining only if there are clouds in the sky." You can probably imagine how these two statements are saying the same thing - whenever it's raining outside, it is a safe conclusion there are clouds in the sky as well. Some other wordings that communicate the same information use either "sufficient" or "necessary." For example, "Raining is a sufficient condition for it to be cloudy," and "Being cloudy is a necessary condition for it to be raining." Here is a table summarizing the different wordings.
The following statements are equivalent:
- If \(p\), then \(q\).
- \(q\) only if \(p\).
- \(p\) is sufficient for \(q\).
- \(q\) is necessary for \(p\).
In everyday life, we often have a stronger meaning in mind when we use a conditional statement. Consider “If you submit your hours today, then you will be paid next Friday.”  The conditional statement if t , then p also includes the inverse of the statement: if not t , then not p . A more compact way to express this statement is “You will be paid next Friday if and only if you submit your timesheet today.” A statement of this form is called a biconditional .
A biconditional is a logical conditional statement in which the hypothesis and conclusion are interchangeable.
A biconditional is written as \(p \leftrightarrow q\) and is translated as " \(p\) if and only if \(q^{\prime \prime}\).
Because a biconditional statement \(p \leftrightarrow q\) is equivalent to \((p \rightarrow q) \wedge(q \rightarrow p),\) we may think of it as a conditional statement combined with its converse: if \(p\), then \(q\) and if \(q\), then \(p\). The double-headed arrow shows that the conditional statement goes from left to right and from right to left. A biconditional is considered true as long as the hypothesis and the conclusion have the same truth value; that is, they are either both true or both false.
\(\begin{array}{|c|c|c|}
\hline p & q & p \leftrightarrow q \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
Notice that the fourth row, where both components are false, is true; if you don’t submit your timesheet and you don’t get paid, the person from payroll told you the truth.
Suppose this statement is true: “The garbage truck comes down my street if and only if it is Thursday morning.” Which of the following statements could be true?

- It is Wednesday at 11:59PM and the garbage truck did not come down my street today.
Solution
- This cannot be true. This is like the second row of the truth table; it is true that I just experienced Thursday morning, but it is false that the garbage truck came.
- This cannot be true.
- This could be true.
Suppose this statement is true: “I wear my running shoes if and only if I am exercising.” Determine whether each of the following statements must be true or false.
- I am exercising and I am not wearing my running shoes.
- I am wearing my running shoes and I am not exercising.
- I am not exercising and I am not wearing my running shoes.
- Answer
-
Choices a & b are false; c is true.
Create a truth table for the statement \((A \vee B) \leftrightarrow \sim C\)
Solution
Whenever we have three component statements, we start by listing all the possible truth value combinations for \(A, B,\) and \(C .\) After creating those three columns, we can create a fourth column for the hypothesis, \(A \vee B\). Now we will temporarily ignore the column for \(C\) and focus on \(A\) and \(B\), writing the truth values for \(A \vee B\).
\(\begin{array}{|c|c|c|}
\hline A & B & C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}\)
\(\begin{array}{|c|c|c|c|}
\hline A & B & C & A \vee B \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}\)
Next we can create a column for the negation of \(C\). (Ignore the \(A \vee B\) column and simply negate the values in the \(C\) column.)
\(\begin{array}{|c|c|c|c|c|}
\hline A & B & C & A \vee B & \sim C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}\)
Finally, we find the truth values of \((A \vee B) \leftrightarrow \sim C\). Remember, a biconditional is true when the truth value of the two parts match, but it is false when the truth values do not match.
\(\begin{array}{|c|c|c|c|c|c|}
\hline A & B & C & A \vee B & \sim C & (A \vee B) \leftrightarrow \sim C \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\end{array}\)
To illustrate this situation, suppose your boss needs you to do either project \(A\) or project \(B\) (or both, if you have the time). If you do one of the projects, you will not get a crummy review ( \(C\) is for crummy). So \((A \vee B) \leftrightarrow \sim C\) means "You will not get a crummy review if and only if you do project \(A\) or project \(B\)." Looking at a few of the rows of the truth table, we can see how this works out. In the first row, \(A, B,\) and \(C\) are all true: you did both projects and got a crummy review, which is not what your boss told you would happen! That is why the final result of the first row is false. In the fourth row, \(A\) is true, \(B\) is false, and \(C\) is false: you did project \(A\) and did not get a crummy review. This is what your boss said would happen, so the final result of this row is true. And in the eighth row, \(A, B\), and \(C\) are all false: you didn't do either project and did not get a crummy review. This is not what your boss said would happen, so the final result of this row is false. (Even though you may be happy that your boss didn't follow through on the threat, the truth table shows that your boss lied about what would happen.)


11.6: Criticism of Colleagues
- Write an internal critical message to another colleague.
As the workplace shifts to a more team-oriented, cross-functional environment, employees are being asked to provide feedback on colleagues on a regular basis. This can be a good thing because personal performance need not rely exclusively on the evaluation of one individual manager. Who better to provide feedback than a fellow teammate that has daily interaction with a colleague?
It does not take long to realize that providing feedback on a coworker could be uncomfortable, especially if there is criticism involved. This would be a situation in which a constructive writing strategy would be useful. Here are some guidelines to ensure that your critical comments are perceived constructively:
- Direct your comments at the behavior and not the person.
- Keep the tone of your message neutral and objective—think about how you would feel reading the same comments about yourself.
- Try to include positive behaviors as well as behavior to be improved.
- Suggest positive steps in resolving the issue.
Here are some examples of constructive criticism addressed to a co-worker.
You are doing a great job making the customer happy and compensating with extra hard work when things get a little behind. You could perhaps spend some time providing more training for the team to take some of the extra work off you and improve their skills in the process. You are fun to work with, especially when you share more of yourself with the team. Your dedication and motivation set a great example for the team. Your intimate knowledge of the technology of the project makes you indispensable. Sometimes when you are asked to help with a problem, you simply create a fix quickly without explaining the solution. It would be very helpful to include the team and explain the issues even if it slows you down a bit. You are very professional and focused on the work running the project. You maintain a calm attitude even when large problems arise. It would be good if you could pay more attention to the tone of your direction to the team when assigning new tasks.Contributors and Attributions
- Criticism of Colleagues. Authored by : Robert Danielson. Provided by : Lumen Learning. License : CC BY: Attribution


1.1: Introduction
Learning Objectives
- Discuss methods for blood sampling
- Compare/contrast management of peripheral venous access devices and central venous access devices
- Review infection control principles associated with peripheral venous access devices
- Explain the principles of peripheral intravenous site selection and contraindications
- Discuss appropriate selection of IV catheter type and size
- Explain nurse management of IV therapy and peripheral IV access devices
- Identify modifications for performing IV therapy across the life span
- Outline nursing management for patients with a patient-controlled analgesia (PCA) pump
- Describe nursing implications for a patient with an epidural infusion for pain management
Intravenous (IV) therapy is an important part of clinical care. It can be used to restore fluids, administer blood products or medications, or serve as an alternate route for nutrition when the gastrointestinal tract is not functioning adequately. IV therapy is a common intervention in nursing practice and useful for rapidly addressing symptoms and restoring hemostasis. Although initiating IV therapy is a common nursing intervention, it is an invasive skill and requires diligent safety practices to prevent and address complications.


6.5: Summary
6.1 Infection Cycle
By understanding the conditions that foster the spread of infection, nurses can implement evidence-based interventions to break the cycle and stop the chain of infection . Nurses must have a thorough understanding of how to protect themselves and their patients from exposure to harmful pathogens. Nurses should gather the signs and symptoms of infections through examination, patient history, and healthcare records. The infection cycle has six specific phases that all must be completed in order for an infection to matriculate. These six pieces are causative agent, reservoir , portal of exit , mode of transmission , portal of entry , and susceptible host . Breaking this chain at any point can stop the spread of infection. Understanding the various types of immune responses will help nurses to identify infection and initiate treatment. The human body’s immune system provides a mechanism for staying healthy through protection against harmful pathogens. An immune response can be classified as either nonspecific, meaning it targets pathogens in a nonspecific manner, or specific, meaning it allows for a high level of adaption and effectiveness against a specific pathogen .
6.2 Asepsis and PPE
Comprehensive knowledge related to infection prevention and control is essential for nurses to protect themselves and patients from the transmission of infection. The two types of asepsis are medical and surgical. Medical asepsis is often referred to as clean technique, while surgical asepsis is often referred to as sterile technique . Understanding the use of personal protective equipment and knowing when to use which type of equipment are key components of infection prevention. By incorporating thorough assessment and applying prevention strategies, nurses are empowered to reduce transmission of pathogens within healthcare environments and the community. One major risk factor to hospital admittance is healthcare-associated infections. The most common HAIs are associated with invasive devices: catheter-associated urinary tract infections (CAUTIs), central line–associated blood stream infections (CLABSIs), ventilator-associated pneumonia (VAP), and surgical site infections (SSIs). These infections often have intensive treatment and compound on the original reason for hospitalization, leading to increased healing time for patients.
6.3 Sterile Technique
Environmental factors play a large role in infection prevention and transmission. Environmental cleaning is a fundamental principle of infection prevention because contaminated equipment and surfaces play a major role in transmission. Improving cleaning practices and increasing compliance reduces the exposure to pathogens and, therefore, reduces incidences of infection. To minimize the risk of transmission of infection, there are three levels of cleaning of which nurses should be knowledgeable: sanitization, disinfection, and sterilization. Moreover, it is critical that nurses understand and utilize the principles of the skill of sterile technique as an essential element of patient safety.
6.4 Infection Control and Patient Safety
Nurses play a critically important role in ensuring patient safety while providing care directly to patients. When standard precautions are not enough to protect patients and healthcare workers against pathogens, nurses must use transmission-based precautions, which provide additional support for infection prevention. Nurses can advocate for themselves, patients, and families to ensure that agencies follow proper policies and procedures to promote the highest patient outcomes. All healthcare employees must receive proper training and education regarding how to reduce their risk for exposure. This may include exposure to certain substances as well as sharps safety. Emphasis must be placed on following proper hand hygiene , and proper PPE protocols can greatly reduce risk of exposure. Special considerations should be made regarding the use of antibiotics to not increase the incidence of MDROs. Patient education on this topic helps to prevent infection with these resistant bacterial agents.


5: Module 5 - Insurance and Bankruptcy
-
- 5.2: Bankruptcy
- Bankruptcy is as an aspect of financing, a system that permits creditors to receive an equitable distribution of the bankrupt person’s assets and promises new hope to debtors facing impossible financial burdens. Without such a law, we may reasonably suppose that the level of economic activity would be far less than it is, for few would be willing to risk being personally burdened forever by crushing debt. Bankruptcy gives the honest debtor a fresh start and resolves disputes among creditors.
-
- 5.2.1: Introduction to Bankruptcy and Overview of the 2005 Bankruptcy Act
- 5.2.2: Case Administration; Creditors’ Claims; Debtors’ Exemptions and Dischargeable Debts; Debtor’s Estate
- 5.2.3: Chapter 7 Liquidation
- 5.2.4: Chapter 11 and Chapter 13 Bankruptcies
- 5.2.5: Alternatives to Bankruptcy
- 5.2.6: Cases
- 5.2.7: Summary and Exercises


16.3A: Types of WBCs
The different types of white blood cells (leukocytes) include neutrophils, basophils, eosinophils, lymphocytes, monocytes, and macrophages.
- Distinguish between the two major types of leukocytes (white blood cells): granulocytes and agranulocytes
Key Points
- The two main types of leukocytes are granulocytes and mononuclear leukocytes (agranulocytes).
- Leukocytes arise from hemopoietic stem cells in the bone marrow.
- Leukocytes are involved in pathogen recognition, phagocytosis (ingestion of particles), pathogen destruction, inflammation mediation, and antigen presentation.
- Granulocytes include neutrophils, basophils, eosinophils, and mast cells. Their granules contain enzymes that damage or digest pathogens and release inflammatory mediators into the bloodstream.
- Mononuclear leukocytes include lymphocytes, monocytes, macrophages, and dendritic cells. This group is involved in both innate and adaptive immune system function.
Key Terms
- endocytosed : Engulfed during the process by which the plasma membrane of a cell folds inwards to ingest material.
- antigen : A substance, usually foreign, that induces an immune response.
- pathogen : Any organism or substance, especially a microorganism, capable of causing disease. Examples include bacteria, viruses, protozoa, or fungi. Microorganisms are not considered pathogenic until the population has grown large enough to cause disease.
White blood cells (WBCs), or leukocytes, are immune system cells that defend the body against infectious disease and foreign materials. There are several different types of WBCs. They share commonalities but are distinct in form and function. WBCs are produced in the bone marrow by hemopoeitic stem cells, which differentiate into either lymphoid or myeloid progenitor cells. A major distinguishing feature is the presence of granules; white blood cells are often characterized as granulocytes or agranulocytes.
Granulocytes
Granulocytes, also known as polymorphonuclear (PMN) leukocytes, are characterized by stained granules within their cytoplasm under a microscope. These granules are membrane-bound enzymes that act primarily in the digestion of endocytosed particles. They may also cause granule dependent cell-mediated apoptosis through the release of perforins, granzymes, and proteases. The nucleus contains multiple lobes (polymorphonuclear) as opposed to a single rounded lobe. Granulocytes contain toll-like receptors that allow them to recognize pathogen-associated molecular patterns (PAMPS). All categories except neutrophils contain IgE receptors that implicate them in allergic responses. There are four types of granulocytes:
Granulocytes : From left to right, a neutrophil, an eosinophil, and a basophil.
- Neutrophils defend against bacterial or fungal infection and other very small inflammatory processes. They are usually the first responders to microbial infection. Their activity and death in large numbers from degranulation forms purulent necrosis (pus).
- Eosinophils primarily deal with parasitic infections. They are also the predominant inflammatory cells in allergic reactions.
- Basophils are chiefly responsible for short-term inflammatory response (particularly from allergy or irritation) by releasing the chemical histamine, which causes the vasodilation that occurs with inflammation.
- Mast cells function similarly to basophils in that they often mediate inflammation, but are more common and arise from a different hemopoeitic lineage.
Mononuclear Leukocytes
Mononuclear (MN) leukocytes are characterized by a single round nucleus within the cytoplasm. Some MN leukocytes contain granules while others do not, but the members of this group are sometimes considered agranulocytes by naming convention. MN leukocytes contain lysosomes, small vesicles containing digestive enzymes that break down foreign matter that is endocytosed by the cell during phagocytosis. The cells include:
- Lymphocytes, which come in three types. B-lymphocytes produce antibodies in the humoral immune response. T-lymphocytes participate in the cell-mediated immune response. NK cells are cytotoxic cells that participate in the innate immune response by killing virally infected and tumor cells and mediating fever and long-lasting inflammation. B and T lymphocytes contain MHC antigen receptors and their activity is antigen-specific. Other leukocytes will attack any pathogen but cannot distinguish between different types of pathogens.
- Monocytes are large leukocytes that differentiate into macrophages and dendritic cells under varying conditions, while performing similar functions in phagocytosis and antigen presentation (the process by which molecular components are presented to lymphocytes to stimulate an adaptive immune response). Monocytes and their progeny contain toll-like receptors and granules.
- Macrophages are monocytes that have migrated out of the blood stream and into the internal body tissues. They destroy necrotic cell debris and foreign material including viruses and bacteria, and can present antigens to naive lymphocytes. They typically arrive at the site of inflammation one to three days after the initial neutrophil response to clean up dead neutrophils, cellular debris, and remaining pathogens.
- Dendritic cells are monocytes that have migrated to cells that are in contact with the external environment, such as the skin, intestines, or respiratory epithelium. Their name comes branched projections called dendrites, which increase their surface area. They phagocytize pathogens and present antigens to naive lymphocytes.
A Macrophage : A macrophage phagocytizes two smaller particles, possibly pathogens


2.4: The “Role” of Variables- Predictors and Outcomes
Okay, I’ve got one last piece of terminology that I need to explain to you before moving away from variables. Normally, when we do some research we end up with lots of different variables. Then, when we analyse our data we usually try to explain some of the variables in terms of some of the other variables. It’s important to keep the two roles “thing doing the explaining” and “thing being explained” distinct. So let’s be clear about this now. Firstly, we might as well get used to the idea of using mathematical symbols to describe variables, since it’s going to happen over and over again. Let’s denote the “to be explained” variable Y , and denote the variables “doing the explaining” as X1, X2, etc.
Now, when we doing an analysis, we have different names for X and Y , since they play different roles in the analysis. The classical names for these roles are independent variable (IV) and dependent variable (DV). The IV is the variable that you use to do the explaining (i.e., X) and the DV is the variable being explained (i.e., Y ). The logic behind these names goes like this: if there really is a relationship between X and Y then we can say that Y depends on X, and if we have designed our study “properly” then X isn’t dependent on anything else. However, I personally find those names horrible: they’re hard to remember and they’re highly misleading, because (a) the IV is never actually “independent of everything else” and (b) if there’s no relationship, then the DV doesn’t actually depend on the IV. And in fact, because I’m not the only person who thinks that IV and DV are just awful names, there are a number of alternatives that I find more appealing. The terms that I’ll use in these notes are predictors and outcomes. The idea here is that what you’re trying to do is use X (the predictors) to make guesses about Y (the outcomes). 4 This is summarised in Table 2.2.
4 Annoyingly, though, there’s a lot of different names used out there. I won’t list all of them – there would be no point in doing that – other than to note that R often uses “response variable” where I’ve used “outcome”, and a traditionalist would use “dependent variable”. Sigh. This sort of terminological confusion is very common, I’m afraid.


15.8: Excel Table in PowerPoint
Learning Objectives
- Create PowerPoint slide with integrated Excel table.
Cameron has an Excel file with a pertinent table for the sales presentation. He selects slide 3, types in a slide title, and then follows these steps to insert the Excel table into PowerPoint.
Option 1 for Table Integration
In the text box, Cameron hovers over the icons in the middle and clicks on the Insert Table icon.
He doesn’t have an Excel table in a file to copy into his PowerPoint presentation, so he creates one through the following steps:
- Insert Table Box. This box appears where Cameron decides the size of his table.
- Number of Columns. Clicks the down arrow until the number is 4 columns.
- Number of Rows. Clicks the up arrow until the number is 6 rows.
- OK Button. This button creates a little Excel table within Cameron’s PowerPoint presentation slide.
The Excel table is now in his PowerPoint presentation and ready for data to be entered. Cameron now enters his information into the table.
Note: If at any time Cameron wants to change the color of the table he can click on the Table styles in the bar above.
Cameron types in the sales figures from the three different stores, and wants to make additional changes. He has already bolded the title rows (using the Bold button in the Home Tab), but the type is difficult to read from a distance.
- He selects the table title.
- He right-clicks to bring up the font size and selects size 40.
- He selects the remaining rows and uses the upper font bar to change the font size to 20pt.
- He decides to change the title by clicking within the title box, then typing in a new title.
Cameron steps back and looks at the table. He notices the table is not centered on the slide, so he clicks on the table and moves his cursor to the edge of the table until he sees the four arrow symbol. He drags the whole table to the center of the slide.
He now makes a note under the table by clicking on the Insert tab and the Text Box button.
He puts his cursor under the table and begins to type his note into the slide. This step is now complete.
Cameron reads through the slide to make sure it is clear and decides to change the table title. He clicks in the table title and retypes the words to say “Q3 Campaign.” Now the slide is complete, and he is ready for another slide.
Option 2 for Table Integration
What if Cameron has an existing sales campaign Excel table he can use? With this scenario, he copies it directly into his PowerPoint presentation so there’s no need to make his own table. Here is how Cameron inserts an existing Excel file table:
- Open Excel application. Cameron has the RR Sales Campaign Excel file open on his computer.
- Select the table. Cameron selects the entire table and clicks with the right mouse button.
- Copy the Excel table. Next, Cameron opens the selects the Copy function. Cameron can also use the Ctrl+C keys to copy the table.
- Select PowerPoint . Cameron clicks on the PowerPoint window to bring it forward (he could also toggle Alt+Tab if the window is behind the Excel window).
- Select table area. Cameron clicks with the right mouse button in the Table box.
- Paste Excel table. Cameron selects the Paste function in the menu and the table is placed in the slide (He can also use the Ctrl+V keys to paste in the table).
He can now resize, move, center, bold, add the text note, and change the font of the table like in the Option 1 example to optimally display the table on the slide. The table slide is now set, and Cameron is ready for an additional slide with a chart indicating the campaign funding divisions between stores.
Contributors and Attributions
- Excel Table in PowerPoint. Authored by : Sherri Pendleton. Provided by : Lumen Learning. License : CC BY: Attribution


8.3: Cohen's Standards for Small, Medium, and Large Effect Sizes
Cohen's d is a measure of "effect size" based on the differences between two means. Cohen’s \(d\), named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.
| Size of effect | \(d\) |
|---|---|
| Small | ≤ 0.2 |
| Medium | 0.2 - 0.8 |
| Large | ≥ 0.8 |
Table \(\PageIndex{1}\) Cohen's Standard Effect Sizes
Cohen's \(d\) is the measure of the standardized difference between two means, calculated as follows: \[d=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{2}+\frac{s_{2}^{2}}{2}}}\nonumber\]
It is important to note that Cohen's \(d\) does not provide a level of confidence as to the magnitude of the size of the effect comparable to the other tests of hypothesis we have studied. The sizes of the effects are simply indicative.
Example \(\PageIndex{1}\)
Calculate Cohen’s \(d\) for the difference between two groups, where \(\overline{x}_{1}=4\), \(s_{1}=1.5\), \(n_{1}=11\), and \(\overline{x}_{2}=3.5\), \(\mathrm{s}_{2}=1\), \(n_{2}=9\). Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.
- Answer
-
\(d = 0.89\)
The effect is large because 0.89 is greater than Cohen’s value of 0.8 for large effect sizes. The size of the differences of the means for the two companies is large, indicating that there is a meaningful difference between them.


3.3: How It’s All Connected
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The snail vignette illustrates how teachers can help children make connections and thereby make meaning. This exploration allowed children to investigate and learn about creatures from the outdoor environment in the classroom. In doing so, the children were able to make meaning about snails’ natural habitats while encountering opportunities to engage in integrated learning in every domain.
Young children’s experiences at home and in their communities are a powerful source of connections. Teachers nurture children’s appetites for learning and making meaning by building upon the knowledge children bring to the preschool setting. For example, children may come to preschool with knowledge of many family stories. Their teachers may have observed that the children used the stories in the dramatic play area. However, the children did not seem to be aware that their stories could be written down and then read by someone else. In such a case, teachers can partner with families to create a story dictation study. In planning the snail exploration, the teachers and family members may ask:
Teachers can explore these questions and see where the exploration leads. [1]
[1] California Preschool Curriculum Framework, Volume 1 by the California Department of Education is used with permission (pg. 23-24)


7.19G: Small Regulatory RNAs
Proteolytic degradation, or proteolysis, is a key factor that controls protein concentration and function.
LEARNING OBJECTIVES
Describe protein degradation
Key Takeaways
The Process of Protein Degradation in a Proteosome : Schematic of the proteolytic degradation pathway that utilizes proteasomal complexes. The protein is tagged with several ubiquitin signals that target the proteasome. Once the protein arrives at the proteasome, the protein is degraded into its amino acids which are then reused for synthesis of new proteins.
Proteolytic degradation is necessary in the regulation of cellular processes and function. The breakdown of proteins into smaller polypeptides, or its respective amino acids, are necessary for metabolic and cellular homeostasis. Polypeptides are commonly broken down via hydrolysis of the peptide bonds by utilizing a class of enzymes called proteases . However, proteolytic degradation can also occur utilizing various mechanisms, including intramolecular digestion and non-enzymatic methods. The mechanisms of proteolytic degradation are necessary for obtaining amino acids via degradation of digested proteins, preventing accumulation or abnormal concentrations of proteins, and by regulating cellular processes by removing proteins no longer needed.
Proteasomes are protein complexes that function in the degradation of unneeded or damaged proteins via proteolysis. The proteasomes are a major component of a complex and highly regulated mechanism. The proteasome is able to degrade proteins based on the presence of a ubiquitin protein. This ubiquitin sequence is a modification to proteins that are targeted for degradation. The recognition of this ubiquitin signal by the proteasome results in degradation of the protein into its amino acids, which are then recycled and reused for the synthesis of new proteins. The proteasomal degradation pathway is the major pathway utilized to ensure proteolytic degradation. It is necessary for homeostasis functioning in controlling cell cycle and gene expression, for example.
In addition to proteasomal complexes, the organelle, the lysosomes are also used to ensure protein degradation. The intracellular process that utilizes lysosomes involves autophagy. The lysosomal pathway, in comparison to the proteasomal pathway, is typically non-selective. The lysosome contains proteases that are able to target and degrade proteins.
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
- Bacterial small RNA. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Bacterial_small_RNA . License : CC BY-SA: Attribution-ShareAlike
- Regulation of gene expression. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Regulat...ene_expression . License : CC BY-SA: Attribution-ShareAlike
- Antisense RNA. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Antisense_RNA . License : CC BY-SA: Attribution-ShareAlike
- Hok/sok system. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Hok/sok_system . License : CC BY-SA: Attribution-ShareAlike
- morphogenesis. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/morphogenesis . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- Attenuator (genetics). Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Attenuator_(genetics) . License : CC BY-SA: Attribution-ShareAlike
- operons. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/operons . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- Riboswitch. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Riboswi...rld_hypothesis . License : CC BY-SA: Attribution-ShareAlike
- Riboswitches. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Riboswitches . License : CC BY-SA: Attribution-ShareAlike
- aptamer. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/aptamer . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- TPP riboswitch pdb-2hoj. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:TP...h_pdb-2hoj.png . License : CC BY-SA: Attribution-ShareAlike
- RpoS. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/RpoS . License : CC BY-SA: Attribution-ShareAlike
- Sigma factor. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sigma_factor . License : CC BY-SA: Attribution-ShareAlike
- Anti-sigma factors. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Anti-sigma_factors . License : CC BY-SA: Attribution-ShareAlike
- Bacterial small RNA. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Bacterial_small_RNA . License : CC BY-SA: Attribution-ShareAlike
- growth phase transitions. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/growth%...%20transitions . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- TPP riboswitch pdb-2hoj. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:TP...h_pdb-2hoj.png . License : CC BY-SA: Attribution-ShareAlike
- PDB 1h3l EBI. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...B_1h3l_EBI.jpg . License : Public Domain: No Known Copyright
- RpoS. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/RpoS%23...ontrol_of_rpoS . License : CC BY-SA: Attribution-ShareAlike
- sigma factor. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/sigma%20factor . License : CC BY-SA: Attribution-ShareAlike
- RpoS protein. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/RpoS%20protein . License : CC BY-SA: Attribution-ShareAlike
- oxidative stress. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/oxidative_stress . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- TPP riboswitch pdb-2hoj. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:TP...h_pdb-2hoj.png . License : CC BY-SA: Attribution-ShareAlike
- PDB 1h3l EBI. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...B_1h3l_EBI.jpg . License : Public Domain: No Known Copyright
- Image:Ta7.jpg - OpenWetWare. Provided by : Open Wetware. Located at : http://openwetware.org/wiki/Image:Ta7.jpg . License : CC BY-SA: Attribution-ShareAlike
- Proteolysis. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Proteol...in_degradation . License : CC BY-SA: Attribution-ShareAlike
- Proteasome. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Proteasome%23Proteolysis . License : CC BY-SA: Attribution-ShareAlike
- ubiquitin. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/ubiquitin . License : CC BY-SA: Attribution-ShareAlike
- proteases. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/proteases . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- TPP riboswitch pdb-2hoj. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:TP...h_pdb-2hoj.png . License : CC BY-SA: Attribution-ShareAlike
- PDB 1h3l EBI. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...B_1h3l_EBI.jpg . License : Public Domain: No Known Copyright
- Image:Ta7.jpg - OpenWetWare. Provided by : Open Wetware. Located at : http://openwetware.org/wiki/Image:Ta7.jpg . License : CC BY-SA: Attribution-ShareAlike
- Ubiquitylation. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...uitylation.png . License : CC BY-SA: Attribution-ShareAlike
- Housekeeping gene. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Housekeeping_gene . License : CC BY-SA: Attribution-ShareAlike
- Antisense RNA. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Antisense_RNA . License : CC BY-SA: Attribution-ShareAlike
- Bacterial small RNA. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Bacterial_small_RNA . License : CC BY-SA: Attribution-ShareAlike
- RNA polymerase. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/RNA%20polymerase . License : CC BY-SA: Attribution-ShareAlike
- Hok sok system R1 plasmid present. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ho...id_present.gif . License : Public Domain: No Known Copyright
- File:Trp operon attenuation.svg - Wikipedia, the free encyclopedia. Provided by : Wikipedia. Located at : en.Wikipedia.org/w/index.php?...ion.svg&page=1 . License : CC BY-SA: Attribution-ShareAlike
- TPP riboswitch pdb-2hoj. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:TP...h_pdb-2hoj.png . License : CC BY-SA: Attribution-ShareAlike
- PDB 1h3l EBI. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...B_1h3l_EBI.jpg . License : Public Domain: No Known Copyright
- Image:Ta7.jpg - OpenWetWare. Provided by : Open Wetware. Located at : http://openwetware.org/wiki/Image:Ta7.jpg . License : CC BY-SA: Attribution-ShareAlike
- Ubiquitylation. Provided by : Wikimedia. Located at : commons.wikimedia.org/wiki/Fi...uitylation.png . License : CC BY-SA: Attribution-ShareAlike
- Antisense DNA oligonucleotide. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:An...nucleotide.png . License : CC BY: Attribution
Key Points
- The major mechanism of proteolytic degradation utilized by the cell, is via the proteasomal pathway. Proteins that are degraded via the proteasomal complex are tagged via the addition of a ubiquitin signal.
- An additional mechanism utilized for proteolytic degradation is via the lysosomal pathway. The lysosome contains proteases which target proteins for degradation.
- Proteolysis is necessary to control protein concentration and prevent abnormal accumulation.
- Upon protein degradation, the amino acids are typically reused and recycled for the synthesis of new proteins.
Key Terms
- ubiquitin : A small regulatory protein sequence that directs proteins to specific compartments within the cell. Specifically, a ubiquitin tag directs the protein to a proteasome, which destroys and recycles the components.
- proteases : A class of enzymes that can cleave proteins.


1.1: Functions and Function Notation
- Refresher Topics
-
-
Sets and Numbers
- Sets and Set Notation (definition and roster notation only)
-
Factoring Techniques
- Greatest Common Factor
- Factoring Trinomials (of the form \( x^2 + bx + c \))
-
Solving Equations
- Solving Quadratic Equations (by factoring and Extraction of Roots only)
- Solving Radical Equations (single square root only)
-
An Overview of Functions
- Relations and Mappings
-
Sets and Numbers
- Foundational Topics
-
-
Sets and Numbers
- Sets of Numbers (all but the complex numbers)
- Properties of Real Numbers
-
Arithmetic
- Mathematical Operations - Addition, Subtraction, Multiplication, and Division
- Mathematical Operations - Powers and Square Roots
- The Order of Operations
- Simplifying Fractions
- Mathematical Operations - Higher-Index Roots
- The Absolute Value of a Number
-
Simplifying Expressions
- Multiplying and Distributing Polynomial Expressions
- Simplifying Ratios of Polynomials
-
Solving Equations
- Solving Linear Equations
-
Graphs
- The Cartesian Coordinate System
- Ordered Pairs and Graphing Relations by Point-Plotting
-
Sets and Numbers
- Determine whether a relation represents a function.
- Find the value of a function.
- Determine whether a function is one-to-one.
- Use the Vertical Line Test to identify functions.
- Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set of the first components of each ordered pair is called the domain and the set of the second components of each ordered pair is called the range .
Consider the following set of ordered pairs.\[ \{ (1,2), (2,4), (3,6), (4,8), (5,10) \} \nonumber \]The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first. The domain is\[ \{1, 2, 3, 4, 5\} \nonumber \]and the range is\[ \{2, 4, 6, 8, 10\} \nonumber \]Note that each value in the domain is also known as an input value , and each value in the range is also known as an output value .
When referring to an arbitrary element from the set of input values, we often use the variable \( x \). In such cases, we call \( x \) the independent variable . Likewise, the variable \( y \) is often used when referencing an arbitrary element from the set of output values, and \( y \) is called the dependent variable .
While \( x \) and \( y \) are traditionally used in lower-level math courses as the independent and dependent variables, respectively, this need not always be the case. We can use any letter (or picture, for that matter) to represent these variables. The choice is literally limitless.
A function is a relation that assigns a single value in the range to each value in the domain. In a very real sense, functions are a special subclass of relations. While a relation can relate the input of \( 5 \) to the output \( 7 \) and then the input of \( 5 \) to the output of \( 10 \), functions are not allowed such silliness.
For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, \(\{1,2,3,4,5\}\), is paired with exactly one element in the range, \(\{2,4,6,8,10\}\).
Now, let’s consider the set of ordered pairs that relates the terms "even" and "odd" to the first five natural numbers. It would appear as\[ \{(\text{odd},1),(\text{even},2),(\text{odd},3),(\text{even},4),(\text{odd},5)\}. \nonumber \]Notice that each element in the domain, \(\{\text{even}, \text{odd} \} \), is not paired with exactly one element in the range, \(\{1,2,3,4,5\}\). For example, the term "odd" corresponds to three values from the range, \(\{1,3,5\}\) and the term "even" corresponds to two values from the range, \(\{2,4\}\). This violates the definition of a function, so this relation is not a function. Figure \( \PageIndex{ 1 } \) compares relations that are functions and not functions.
Figure \( \PageIndex{ 1 } \)
Figure \( \PageIndex{ 1(a) } \) is a function because each input is associated with a single output. Note that inputs \( q \) and \( r \) both give output \( n \). This is perfectly fine. The definition of a function only restricts a single input from leading to two distinctly different outputs. Figure \( \PageIndex{ 1(b) } \) is also a function. In this case, each input is associated with a single output. Finally, Figure \( \PageIndex{ 1(c) } \) is not a function because input \( q \) is associated with two different outputs (\( y \) and \( z \)).
Functions are so important throughout the rest of your mathematical career that we should formalize their definition.
A function is a relation in which each possible input value leads to exactly one output value. We say the output is a function of the input. The set of input values makes up the domain of the function, and the corresponding set of output values makes up the range of the function.
The coffee shop menu, shown in Figure \( \PageIndex{ 2 } \), consists of items and their prices.
- Is price a function of the item?
- Is the item a function of the price?
Figure \( \PageIndex{ 2 } \)
- Solutions
-
- Let’s begin by considering the input as the items on the menu. The output values are then the prices (see Figure \( \PageIndex{ 2 } \)). Each item on the menu has only one price, so the price is a function of the item.
-
Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it (see the figure below).
In a particular math class, the overall percent grade corresponds to a grade point average. Is grade point average a function of the percent grade? Is the percent grade a function of the grade point average? The following table shows a possible rule for assigning grade points.
| Percent grade | 0–56 | 57–61 | 62–66 | 67–71 | 72–77 | 78–86 | 87–91 | 92–100 |
| Grade point average | 0.0 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
- Solution
-
For any percent grade earned, there is an associated grade point average, so the grade point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average.
In the grading system given, there is a range of percent grades that correspond to the same grade point average. For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade point average.
The following table lists the five greatest baseball players of all time in order of rank.
| Player | Rank |
|---|---|
| Babe Ruth | 1 |
| Willie Mays | 2 |
| Ty Cobb | 3 |
| Walter Johnson | 4 |
| Hank Aaron | 5 |
- Is the rank a function of the player name?
- Is the player name a function of the rank?
Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent "height is a function of age," we start by identifying the descriptive variables: \( h \) for height and \( a \) for age. The letters \( f \), \( g \), and \( h \) are often used to represent functions just as we use \( x \), \( y \), and \( z \) to represent numbers and \( A \), \( B \), and \( C \) to represent sets.
We then build a structure for the function notation. Since height is a function of age, we would like our notation to match.\[ \begin{array}{rrcl}
& \text{height} & \text{is} & \text{a function of age} \\[6pt]
\implies & h & = & f(a) \\[6pt]
\end{array} \nonumber \]Remember, we can use any letter to name the function. The notation \( f(a) \) shows us that \( f \) depends on \( a \). The value \( a \) must be put into the function \( f \) to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example, \( f(a +b) \) means, "first add \(a\) and \(b\), and then input the result into the function \(f\)." The operations must be performed in this order to obtain the correct result.
The notation \( y = f(x) \) defines a function named \( f \). This is read as "\( y \) is a function of \( x \)." The letter \( x \) represents the input value, or independent variable. The letter \( y \), or \( f(x) \), represents the output value, or dependent variable.
Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Assume that the domain does not include leap years.
- Solution
-
The number of days in a month is a function of the name of the month, so if we name the function \( f \), we write\[ \text{days} = f(\text{month}) \nonumber \]or\[ d = f(m). \nonumber \]The name of the month is the input to a "rule" that associates a specific number (the output) with each input.
For example, \( f(\text{March}) = 31 \) because March has 31 days. The notation \( d = f(m) \) reminds us that the number of days, \( d \) (the output), is dependent on the name of the month, \( m \) (the input).
Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs.
A function \(N=f(y)\) gives the number of police officers, \(N\), in a town in year \(y\). What does \(f(2005)=300\) represent?
- Solution
- When we read \(f(2005)=300\), we see that the input year is \(2005\). The value for the output, the number of police officers (\(N\)), is \(300\). Remember, \(N=f(y)\). The statement \(f(2005)=300\) tells us that in the year \(2005\) there were \(300\) police officers in the town.
Use function notation to express the weight of a pig in pounds as a function of its age in days \(d\).
Instead of a notation such as \(y=f(x)\), could we use the same symbol for the output as for the function, such as \(y=y(x)\), meaning "\(y\) is a function of \(x\)?"
Yes, this is often done, especially in applied subjects that use higher math, such as Physics and Engineering. However, in exploring Mathematics itself, we like to maintain a distinction between a function such as \(f\), which is a rule or procedure, and the output \(y\) we get by applying \( f \) to a particular input \( x \). This is why we usually use notation such as \( y=f(x) \), \( P=W(d) \), and so on.
Representing Functions Using Tables
A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship.
The follow table lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function \( f \) where \( D=f(m) \) identifies months by an integer rather than by name.
|
Month number, \( m \)
(input) |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|
Days in month, \( D \)
(output) |
31 | 28 | 31 | 30 | 31 | 30 | 31 | 31 | 30 | 31 | 30 | 31 |
The next table defines a function \( Q=g(n) \). Remember, this notation tells us that \( g \) is the name of the function that takes the input \( n \) and gives the output \( Q \).
| \( n \) | 1 | 2 | 3 | 4 | 5 |
| \( Q \) | 8 | 6 | 7 | 6 | 8 |
Our final table displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.
| Age in years, \( a \) (input) | 5 | 5 | 6 | 7 | 8 | 9 | 10 |
| Height in inches, \( h \) (output) | 40 | 42 | 44 | 47 | 50 | 52 | 54 |
Which table, Table 5.A, Table 5.B, or Table 5.C, represents a function (if any)?
Table 5.A
| Input | Output |
|---|---|
| 2 | 1 |
| 5 | 3 |
| 8 | 6 |
Table 5.B
| Input | Output |
|---|---|
| –3 | 5 |
| 0 | 1 |
| 4 | 5 |
Table 5.C
| Input | Output |
|---|---|
| 1 | 0 |
| 5 | 2 |
| 5 | 4 |
- Solution
-
Table 5.A and Table 5.B define functions. In both, each input value corresponds to exactly one output value. Table 5.C does not define a function because the input value of 5 corresponds to two different output values.
When a table represents a function, corresponding input and output values can also be specified using function notation.
The function represented by Table 5.A can be represented by writing\[ f(2) = 1, f(5) = 3, \text{ and } f(8) = 6. \nonumber \]Similarly, the statements\[ g(-3) = 5, g(0) = 1, \text{ and } g(4) = 5 \nonumber \]represent the function in Table 5.B.
Table 5.C cannot be expressed in a similar way because it does not represent a function.
Does the following table represent a function?
| Input | Output |
|---|---|
| 1 | 10 |
| 2 | 100 |
| 3 | 1000 |
Finding Input and Output Values of a Function
When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.
When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.
Evaluation of Functions in Algebraic Forms
When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function \( f(x)=5−3x^2 \) can be evaluated by squaring the input value, multiplying by \( 3 \), and then subtracting the product from \( 5 \).
Evaluate \( f(x)=x^2+3x−4 \) at:
- 2
- \( a \)
- \( a+h \)
- Now evaluate \( \frac{f(a+h)−f(a)}{h} \)
- Solutions
-
Replace the \( x \) in the function with each specified value.
-
Because the input value is a number, \( 2 \), we can use simple Algebra to simplify.\[ \begin{array}{rclcl}
f(2) & = & (2)^2 + 3(2) - 4 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 4 + 6 - 4 & & \\[6pt]
& = & 6 & & \\[6pt]
\end{array} \nonumber \] - In this case, the input value is a letter so we cannot simplify the answer any further.\[ f(a) = a^2 + 3a - 4 \nonumber \]
-
With an input value of \( a+h \), we must use the distributive property.\[ \begin{array}{rclcl}
f(a + h) & = & (a + h)^2 + 3(a+ h) - 4 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & a^2 + 2ah + h^2 + 3a + 3h - 4 & \quad & \left( \text{distributing} \right) \\[6pt]
\end{array} \nonumber \] -
In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that\[ f(a + h) = a^2 + 2ah + h^2 + 3a + 3h - 4 \nonumber \]and we know that\[ f(a) = a^2 + 3a - 4. \nonumber \]Now we combine the results and simplify.\[ \begin{array}{rclcl}
\dfrac{f(a + h) - f(a)}{h} & = & \dfrac{\left( a^2 + 2ah + h^2 + 3a + 3h - 4 \right) - \left( a^2 + 3a - 4 \right)}{h} & \quad & \left( \text{substituting} \right)\\[6pt]
& = & \dfrac{a^2 + 2ah + h^2 + 3a + 3h - 4 - a^2 - 3a + 4 }{h} & \quad & \left( \text{distributing} \right)\\[6pt]
& = & \dfrac{2ah + h^2 + 3h}{h} & \quad & \left( \text{combining like terms} \right)\\[6pt]
& = & \dfrac{h\left(2a + h + 3\right)}{h} & \quad & \left( \text{factoring out the GCF} \right)\\[6pt]
& = & \dfrac{\cancelto{1}{h}\left(2a + h + 3\right)}{\cancelto{1}{h}} & \quad & \left( \text{canceling like factors} \right)\\[6pt]
& = & 2a + h + 3 & & \\[6pt]
\end{array} \nonumber \]
-
Because the input value is a number, \( 2 \), we can use simple Algebra to simplify.\[ \begin{array}{rclcl}
Given the function \( h(p)=p^2+2p \), evaluate \( h(4) \).
- Solution
-
To evaluate \( h(4) \), we substitute the value \(4\) for the input variable \( p \) in the given function.\[ \begin{array}{rclcl}
h(4) & = & (4)^2 + 2(4) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 16 + 8 & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 24 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Therefore, for an input of \( 4 \), we have an output of \( 24 \).
Given the function \( g(m)=\sqrt{m - 4} \), evaluate \( g(5) \).
Given the function \( h(p)=p^2+2p \), solve \( h(p)=3 \).
- Solution
-
\[ \begin{array}{rrclcl}
& h(p) & = & 3 & & \\[6pt]
\implies & p^2 + 2p & = & 3 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & p^2 + 2p - 3 & = & 0 & \quad & \left( \text{subtracting }3\text{ from both sides} \right) \\[6pt]
\implies & (p + 3)(p - 1) & = & 0 & \quad & \left( \text{factoring} \right) \\[6pt]
\end{array} \nonumber \]If \( (p+3)(p−1)=0 \), either \( (p+3)=0 \) or \( (p−1)=0 \) (or both of them equal \( 0 \)). We will set each factor equal to \( 0 \) and solve for \( p \) in each case (recall from Elementary Algebra, this is called the Zero Factor Property ).\[ \begin{array}{rcl}
(p + 3) = 0 & \implies & p = -3 \\[6pt]
(p - 1) = 0 & \implies & p = 1 \\[6pt]
\end{array} \nonumber \]This gives us two solutions, so we know that \( h(p)=3 \) when the input is either \( p=1 \) or \( p=−3 \). We can also verify by graphing (see the figure below). The graph verifies that \( h(1)=h(−3)=3 \) and (for Example \( \PageIndex{ 7 } \)) \( h(4)=24 \).
Given the function \( g(m)=\sqrt{m-4} \), solve \( g(m)=2 \).
Evaluating Functions Expressed in Formulas
Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation \( 2n+6p=12 \) expresses a functional relationship between \( n \) and \( p \). We can rewrite it to decide if \( p \) is a function of \( n \).
Express the relationship \( 2n+6p=12 \) as a function \( p=f(n) \), if possible.
- Solution
-
To express the relationship in this form, we need to be able to write the relationship where \( p \) is a function of \( n \), which means writing it as\[ p = [\text{expression involving } n]. \nonumber \]\[ \begin{array}{rrclcl}
& 2n + 6p & = & 12 & & \\[6pt]
\implies & 6p & = & 12 - 2n & \quad & \left( \text{subtracting }2n\text{ from both sides} \right) \\[6pt]
\implies & p & = & \dfrac{12 - 2n}{6} & \quad & \left( \text{dividing both sides by }6 \right) \\[6pt]
\implies & p & = & \dfrac{12}{6} - \dfrac{2n}{6} & & \\[6pt]
\implies & p & = & 2 - \dfrac{n}{3} & \quad & \left( \text{canceling like factors} \right) \\[6pt]
\end{array} \nonumber \]Therefore, \( p \) as a function of \( n \) is written as\[ p = f(n) = 2 - \dfrac{1}{3}n. \nonumber \]
It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.
Does the equation \( x^2+y^2=1 \) represent a function with \( x \) as input and \( y \) as output? If so, express the relationship as a function \( y=f(x) \).
- Solution
-
First we subtract \( x^2 \) from both sides, to get\[ y^2 = 1 - x^2. \nonumber \]We now try to solve for \( y \) in this equation.\[ \begin{array}{rrclcl}
& y^2 & = & 1 - x^2 & & \\[6pt]
\implies & y & = & \pm \sqrt{1 - x^2} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt]
\end{array} \nonumber \]Thus, either \( y = \sqrt{1 - x^2} \) or \( y = -\sqrt{1 - x^2} \). That is, we get two outputs corresponding to the same input. Therefore, this relationship cannot be represented as a single function \( y=f(x) \).
If \( x−8 y^3=0 \), express \( y \) as a function of \( x \).
Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula?
Yes, this can happen. For example, given the equation \( x=y+2^y \), if we want to express \( y \) as a function of \( x \), there is no simple algebraic formula involving only \( x \) that equals \( y \). However, each \( x \) does determine a unique value for \( y \), and there are mathematical procedures by which \( y \) can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for \( y \) as a function of \( x \), even though the formula cannot be written explicitly.
Evaluating a Function Given in Tabular Form
As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.
The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table.
| Pet | Memory span in hours |
|---|---|
| Puppy | 0.008 |
| Adult dog | 0.083 |
| Cat | 16 |
| Goldfish | 2160 |
| Beta fish | 3600 |
At times, evaluating a function in table form may be more useful than using equations. For the pet-memory span information, let's call the function \( P \). The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function \( P \) at the input value of "goldfish." We would write \( P(\text{goldfish})=2160 \). Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function \( P \) seems ideally suited to this function, more so than writing it in paragraph or function form.
Use the provided table to answer the following questions.
- Evaluate \( g(3) \).
- Solve \( g(n)=6 \).
| \( n \) | 1 | 2 | 3 | 4 | 5 |
| \( g \) | 8 | 6 | 7 | 6 | 8 |
- Solutions
-
- Evaluating \( g(3) \) means determining the output value of the function \( g \) for the input value of \( n=3 \). The table output value corresponding to \( n=3 \) is 7, so \( g(3)=7 \).
- Solving \( g(n)=6 \) means identifying the input values, \( n \), that produce an output value of 6. The table shows two solutions: 2 and 4. That is, when we input 2 into the function \( g \), our output is 6. When we input 4 into the function \( g \), our output is also 6.
Using the table from Example \( \PageIndex{ 11 } \), evaluate \( g(1) \).
Finding Function Values from a Graph
Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).
Use the graph below to answer the following questions.
- Evaluate \( f(2) \).
- Solve \( f(x) = 4 \).
- Solutions
-
-
To evaluate \( f(2) \), locate the point on the curve where \( x=2 \), then read the \( y \)-coordinate of that point. The point has coordinates \( (2,1) \), so \( f(2)=1 \) (see the following figure).
-
To solve \( f(x)=4 \), we find the output value 4 on the vertical axis. Moving horizontally along the line \( y=4 \), we locate two points of the curve with output value 4: \( (−1,4) \) and \( (3,4) \). These points represent the two solutions to \( f(x)=4 \): -1 or 3. This means \( f(−1)=4 \) and \( f(3)=4 \), or when the input is -1 or 3, the output is 4 (see the figure below).
-
To evaluate \( f(2) \), locate the point on the curve where \( x=2 \), then read the \( y \)-coordinate of that point. The point has coordinates \( (2,1) \), so \( f(2)=1 \) (see the following figure).
Use the graph from Example \( \PageIndex{ 12 } \) to solve \( f(x) = 1 \).
Determining Whether a Function is One-to-One
Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in the figure at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000.
However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions . As an example, consider a school that uses only letter grades and decimal equivalents, as listed in the following table.
| Letter grade | Grade point average |
|---|---|
| A | 4.0 |
| B | 3.0 |
| C | 2.0 |
| D | 1.0 |
This grading system represents a one-to-one function because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.
To visualize this concept, let’s look again at the two simple functions sketched in Figure \( \PageIndex{ 1(a) } \) and Figure \( \PageIndex{ 1(b) } \). The function in part (a) shows a relationship that is not a one-to-one function because inputs \( q \) and \( r \) both give output \( n \). The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.
A one-to-one function is a function in which each output value corresponds to exactly one input value.
Is the area of a circle a function of its radius? If yes, is the function one-to-one?
- Solution
-
A circle of radius \( r \) has a unique area measure given by \( A = \pi r^2 \), so for any input, \( r \), there is only one output, \( A \). The area is a function of radius \( r \).
If the function is one-to-one, the output value (the area) must correspond to a unique input value (the radius). Any area measure \( A \) is given by the formula \( A= \pi r^2 \). Because areas and radii are positive numbers, there is exactly one solution: \( \sqrt{\frac{A}{\pi}} \). So the area of a circle is a one-to-one function of the circle’s radius.
- Is a balance a function of the bank account number?
- Is a bank account number a function of the balance?
- Is a balance a one-to-one function of the bank account number?
Evaluate the following:
- If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade?
- If so, is the function one-to-one?
Using the Vertical Line Test
As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.
The most common graphs name the input value \( x \) and the output value \( y \), and we say \( y \) is a function of \( x \), or \( y=f(x) \) when the function is named \( f \). The graph of the function is the set of all points \( (x,y) \) in the plane that satisfies the equation \( y=f(x) \). If the function is defined for only a few input values, then the graph of the function is only a few points, where the \( x \)-coordinate of each point is an input value and the \( y \)-coordinate of each point is the corresponding output value. For example, the black dots on the graph in the following figure tell us that \( f(0)=2 \) and \( f(6)=1 \). However, the set of all points \( (x,y) \) satisfying \( y=f(x) \) is a curve. The curve shown includes \( (0,2) \) and \( (6,1) \) because the curve passes through those points.
The Vertical Line Test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value (see the following figure).
Which of the graphs in the following figure represent(s) a function \( y=f(x) \)?
- Solutions
-
If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of the figure. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most \( x \)-values, a vertical line would intersect the graph at more than one point, as shown in the figure below.
Does the graph in the following figure represent a function?
Using the Horizontal Line Test
Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the Horizontal Line Test . Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.
Consider the functions shown in parts (a) and (b) of Example \( \PageIndex{ 14 } \). Are either of the functions one-to-one?
- Solution
-
The function in part (a) is not one-to-one. The horizontal line shown in the following figure intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)
The function in part (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.
Is the graph shown in part (c) of Example \( \PageIndex{ 14 } \) one-to-one?
Identifying Basic Toolkit Functions
In this text, we will be exploring functions - the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do Arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our " toolkit functions ," which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use \( x \) as the input variable and \( y=f(x) \) as the output variable.
We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in the following table.
| Toolkit Functions | ||
|---|---|---|
| Name | Function | Graph |
| Constant |
\( f(x) = c \), where \( c \) is a constant |
|
| Identity |
\( f(x) = x \) |
|
| Absolute value |
\( f(x) = |x| \) |
|
| Quadratic |
\( f(x) = x^2 \) |
|
| Cubic |
\( f(x) = x^3 \) |
|
| Reciprocal |
\( f(x) = \dfrac{1}{x} \) |
|
| Reciprocal squared |
\( f(x) = \dfrac{1}{x^2} \) |
|
| Square root |
\( f(x) = \sqrt{x} \) |
|
| Cube root |
\( f(x) = \sqrt[3]{x} \) |
Access the following online resources for additional instruction and practice with functions.


14: Galvanic Cells
An electrochemical cell is a system in which passage of an electric current through an electrical circuit is linked to an internal cell reaction. A galvanic cell , or voltaic cell, is an electrochemical cell that, when isolated, has an electric potential difference between its terminals; the cell is said to be a seat of electromotive force .
The cell reaction in a galvanic cell differs in a fundamental way from the same reaction (i.e., one with the same reaction equation) taking place in a reaction vessel that is not part of an electrical circuit. In the reaction vessel, the reactants and products are in the same phase or in phases in contact with one another, and the reaction advances in the spontaneous direction until reaction equilibrium is reached. This reaction is the direct reaction .
The galvanic cell, in contrast, is arranged with the reactants physically separated from one another so that the cell reaction can advance only when an electric current passes through the cell. If there is no current, the cell reaction is constrained from taking place. When the electrical circuit is open and the cell is isolated from its surroundings, a state of thermal, mechanical, and transfer equilibrium is rapidly reached. In this state of cell equilibrium or electrochemical equilibrium , however, reaction equilibrium is not necessarily present—that is, if the reactants and products were moved to a reaction vessel at the same activities, there might be spontaneous advancement of the reaction.
As will be shown, measurements of the cell potential of a galvanic cell are capable of yielding precise values of molar reaction quantities of the cell reaction and thermodynamic equilibrium constants, and of mean ionic activity coefficients in electrolyte solutions.
Contributors
- Howard DeVoe, Associate Professor Emeritus, University of Maryland from Thermodynamics and Chemistry


8.1: Acid and Base Neutralization
Learning Objectives
- Write and understand acid-base neutralization reactions.
- Write the neutralization reactions between carboxylic acids and bases.
Neutralization Reactions
This is a reaction that happens when an acid, such as \(\ce{HCl}\), is mixed with a base, such as \(\ce{NaOH}\).
\[\ce{HCl (aq) + NaOH (aq) → NaCl (aq) + H_2O (l)}\nonumber \]
When an acid and a base are combined, water and a salt are the products . Salts are ionic compounds containing a positive ion other than H + and a negative ion other than the hydroxide ion, \(\ce{OH^{-}}\). Double displacement reactions of this type are called neutralization reactions .
When a strong acid and a strong base are combined in the proper amounts, when \([\ce{H^{+}}]\) equals \([\ce{OH^{-}}\)], a neutral solution results in which pH = 7. The acid and base have neutralized each other, and the acidic and basic properties are no longer present.
One of the most concentrated acids in the body is stomach acid , which can be approximated as a 0.05 M hydrochloric acid solution. Special cells in the stomach wall secrete this acid, along with special enzymes, as part of the digestion process. In a laboratory, a 0.05 M solution of hydrochloric acid would dissolve some metals. How does the stomach survive the presence of such a reactive acid?
The stomach has several mechanisms for withstanding this chemical onslaught. First, the lining of the stomach is coated with a thin layer of mucus that contains some bicarbonate ions (HCO 3 − ). These react with the hydrochloric acid to produce water, carbon dioxide, and harmless chloride ions.
HCl(aq) + HCO 3 - (aq) → HOH(l) + CO 2 (g) + Cl - (aq)
If any acid penetrates through the mucus, it can attack the surface layer of stomach cells, called the gastric epithelium . Cells in the gastric epithelium are being constantly shed, so damaged cells are quickly removed and replaced with healthy cells.
However, if the gastric epithelium is destroyed faster than it can be replaced, the acid may reach the wall of the stomach, resulting in ulcers. If an ulcer grows large enough, it can expose blood vessels in the stomach wall, causing bleeding. In extreme situations, the loss of blood through a severe ulcer can threaten a person’s health.
Ulcers can also result from the presence of a certain bacterium— Helicobacter pylori —in the stomach. The mechanism for this ulcer formation is not the same as that for ulcers caused by stomach acid and is not completely understood. However, there are two main treatments for ulcers: (1) antacids to react chemically with excess hydrochloric acid in the stomach and (2) antibiotics to destroy the H. pylori bacteria in the stomach. Some of the common antacids used to neutralize excess stomach acid are milk of magnesia, Mg(OH) 2 ; Tums, CaCO 3 ; and alka-selzer NaHCO 3 .
The balanced double displacement reaction between milk of magnesia is shown below. The hydrochloric acid in the stomach is neutralized by the base in milk of magnesia.
Mg(OH) 2 (s) + 2HCl (aq) → MgCl 2 (aq) + 2HOH(l)
With tums a neutralization reaction shown in equation 1 between stomach acid and calcium carbonate happens. This double displacement reaction produces carbonic acid, H 2 CO 3 which then decomposes to produce carbon dioxide (CO 2 ) and water (HOH) as shown in equation 2.
CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + H 2 CO 3 (aq) Equation 1
H 2 CO 3 (aq) → CO 2 (g) + HOH(l) Equation 2
The two equations may be combined and written as one:
CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + CO 2 (g) + HOH(l)
Salt solutions as a result of neutralization reaction do not always have a pH of 7, however. The ions produced when an acid and base combine may react with the water molecules to produce a solution that is slightly acidic or basic. As a general concept, if a strong acid is mixed with a weak base, the resulting solution will be slightly acidic. If a strong base is mixed with a weak acid, the solution will be slightly basic.
Neutralization of Carboxylic Acids
Carboxylic acids are more acidic than other functional groups in organic chemistry. So carboxylic acids will react with bases such as sodium hydroxide (NaOH), sodium carbonate (Na 2 CO 3 ), and sodium bicarbonate (NaHCO 3 ) to form water and a carboxylic acid salt. Shown below are the neutralization reactions of acetic acid (CH 3 COOH), a component in vinegar.
CH 3 COOH + NaOH(aq) → CH 3 COO − Na + (aq) + H 2 O
2CH 3 COOH + Na 2 CO 3 (aq) → 2CH 3 COO − Na + (aq) + H 2 O + CO 2 (g)
CH 3 COOH + NaHCO 3 (aq) → CH 3 COO − Na + (aq) + H 2 O + CO 2 (g)
In these reactions, the carboxylic acids neutralize basic compounds. With solutions of carbonate and bicarbonate ions, they also form carbon dioxide gas.
Carboxylic acid salts are named in the same manner as inorganic salts: the name of the cation is followed by the name of the organic anion. The name of the anion is obtained by dropping the - ic ending of the acid name and replacing it with the suffix - ate . This rule applies whether we are using common names or International Union of Pure and Applied Chemistry (IUPAC) names:
Example \(\PageIndex{1}\): Propionic Acid + Calcium Hydroxide
Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH 3 CH 2 CO 2 H) with aqueous calcium hydroxide [Ca(OH) 2 ].
Solution
| Steps | Reaction |
|---|---|
|
Write the unbalanced equation. This is a double displacement reaction, so the cations and anions swap to create the water and the salt. |
CH 3 CH 2 CO 2 H(aq) + Ca(OH) 2 (aq)→(CH 3 CH 2 CO 2 ) 2 Ca(aq) + H 2 O(l) |
|
Balance the equation. Because there are two OH − ions in the formula for Ca(OH) 2 , we need two moles of propionic acid, CH 3 CH 2 CO 2 H , to provide H + ions. |
2 CH 3 CH 2 CO 2 H(aq) + Ca(OH) 2 (aq)→(CH 3 CH 2 CO 2 ) 2 Ca(aq) + 2 H 2 O(l) |
Exercise \(\PageIndex{1}\)
Write a balanced chemical equation for the reaction of solid barium hydroxide with dilute acetic acid.
- Answer
-
\[\ce{Ba(OH)2(s) + 2CH3CO2H (aq)→Ba(CH3CO2)2 (aq) + 2H2O(l)} \nonumber\nonumber \]
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
- Deboleena Roy, Ph.D. Dept. of Science and Engineering @ American River College.
-
Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium.
-
Henry Agnew (UC Davis)


16.3: Galvanic Cells
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Learning Objectives
By the end of this section, you will be able to:
- Describe the function of a galvanic cell and its components
- Use cell notation to symbolize the composition and construction of galvanic cells
As demonstration of spontaneous chemical change, Figure \(\PageIndex{1}\): shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.
These observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, Cu 2 + (aq) , which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:
Consider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells , and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells ).
A galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure \(\PageIndex{2}\). The cell is comprised of two half-cells , each containing the redox conjugate pair (“couple”) of a single reactant. The half-cell shown at the left contains the Cu(0)/Cu(II) couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the Ag(I)/Ag(0) couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode . By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell’s reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge . The spontaneous reaction in this cell produces Cu 2 + cations in the anode half-cell and consumes Ag + ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of Cu 2 + in the anode half-cell are balanced by an influx of NO 3 − from the salt bridge, while a flow of Na + into the cathode half-cell compensates for the decreasing Ag + concentration.
Cell Notation
Abbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics , and they are written following a few guidelines:
- The relevant components of each half-cell are represented by their chemical formulas or element symbols
- All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas
- By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode
A verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure \(\PageIndex{2}\): consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:
Consider a different galvanic cell (see Figure \(\PageIndex{3}\)) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:
In this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II) chloride, immersed in which is a platinum cathode. The cell schematic is then written as
Notice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance (Pt) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell’s redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes .
Example \(\PageIndex{1}\): Writing Galvanic Cell Schematics
A galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a 1 M CrCl 3 solution and another in which a copper wire is immersed in 1 M CuCl 2 . Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode half-reaction, the cathode half-reaction, and the overall cell reaction.
Solution
Since the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:
The half-reactions for this cell are
Multiplying to make the number of electrons lost by Cr and gained by Cu 2 + equal yields
Adding the half-reaction equations and simplifying yields an equation for the cell reaction:
Exercise \(\PageIndex{1}\)
Omitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below.
- Answer
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